homework1

Problem 1:

Returns a Lehmer matrix, where

\[ A_{ij} = \begin{cases} i/j, & \text{if }j\geq{i} \\ j/i, & \text{if }j<i \end{cases} \]

lehmer <- function(n = "number of columns and rows, nXn matrix", 
                   roundTo = "rounding of matrix if desired") {
    theVector <- rep(NA, n*n)
    theMatrix <- matrix(theVector, nrow = n, ncol = n)

    for(j in 1:n) {
        for(i in 1:n) {
        if (j >= i) {
            theMatrix[i,j] <- i/j
        }
        else 
            theMatrix[i,j] <- j/i
        }
    }
    
    return(theMatrix)
}

lehmer(n = 6)

##           [,1]      [,2]      [,3]      [,4]      [,5]      [,6]
## [1,] 1.0000000 0.5000000 0.3333333 0.2500000 0.2000000 0.1666667
## [2,] 0.5000000 1.0000000 0.6666667 0.5000000 0.4000000 0.3333333
## [3,] 0.3333333 0.6666667 1.0000000 0.7500000 0.6000000 0.5000000
## [4,] 0.2500000 0.5000000 0.7500000 1.0000000 0.8000000 0.6666667
## [5,] 0.2000000 0.4000000 0.6000000 0.8000000 1.0000000 0.8333333
## [6,] 0.1666667 0.3333333 0.5000000 0.6666667 0.8333333 1.0000000

Problem 2:

Test whether A is symmetric or not.

A <- lehmer(100)
all.equal(t(A), A)

## [1] TRUE

Yes, A is a symmetric matrix.

Problem 3:

Calculate the inverse of A and assign it to ‘C’. Test whether the inverse is correct.

A <- lehmer(10)
C <- solve(A)
round(C %*% A, 3)

##       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
##  [1,]    1    0    0    0    0    0    0    0    0     0
##  [2,]    0    1    0    0    0    0    0    0    0     0
##  [3,]    0    0    1    0    0    0    0    0    0     0
##  [4,]    0    0    0    1    0    0    0    0    0     0
##  [5,]    0    0    0    0    1    0    0    0    0     0
##  [6,]    0    0    0    0    0    1    0    0    0     0
##  [7,]    0    0    0    0    0    0    1    0    0     0
##  [8,]    0    0    0    0    0    0    0    1    0     0
##  [9,]    0    0    0    0    0    0    0    0    1     0
## [10,]    0    0    0    0    0    0    0    0    0     1

Yes, \(C*A\) is the identity matrix.

Problem 4:

Show that AxB does not always equal BxA.

A <- matrix(c(3, 1, 3, -1, 0, -2, 2, 3, -5), nrow = 3)
B <- matrix(c(3, 7, -1, -6, -14, 2, -3, -7, 1), nrow = 3)
# A*B
A %*% B

##      [,1] [,2] [,3]
## [1,]    0    0    0
## [2,]    0    0    0
## [3,]    0    0    0

# B*A
B%*%A

##      [,1] [,2] [,3]
## [1,]   -6    3    3
## [2,]  -14    7    7
## [3,]    2   -1   -1

Problem 5:

A bank makes four kinds of loans to its personal customers, and these loans yield the following annual interest rates to the Bank:

• First mortgage 14%
• Second mortgage 205
• Home improvement 20%
• Personal overdraft 10%

We are interested in the bank’s lending strategy. The information we know is as following:

1. In total $250 million is lent out.
2. First mortgages are 55% of all mortgages (i.e, the first and the second mortgage) issued.
3. Second mortgages are 25% of all the loans issued.
4. The average interest rate on all loans in 15%.

Calculate the lending strategy using matrix inversion.

A <- matrix(c(1,1,1,1,.14,.2,.2,.1,1,0,0,0,0,1,0,0), nrow = 4, byrow = TRUE)
b <- c(250000000, 37500000, 76388889.7, 62500000)
loan_amounts <- solve(A)%*%b
interest_rates <- c(.14, .2, .2, .1)

A

##      [,1] [,2] [,3] [,4]
## [1,] 1.00  1.0  1.0  1.0
## [2,] 0.14  0.2  0.2  0.1
## [3,] 1.00  0.0  0.0  0.0
## [4,] 0.00  1.0  0.0  0.0

b

## [1] 250000000  37500000  76388890  62500000

loan_amounts

##          [,1]
## [1,] 76388890
## [2,] 62500000
## [3,] 31944444
## [4,] 79166666